Ex 7.5, 15

Transcript

Ex 7.5, 15 1﷮ 𝑥﷮4﷯ − 1﷯ Now, 1﷮ 𝑥﷮4﷯ − 1﷯= 1﷮ 𝑥﷮2﷯ + 1﷯ 𝑥﷮2﷯ − 1﷯﷯ = 1﷮ 𝑥﷮2﷯ + 1﷯ 𝑥 − 1﷯ 𝑥 + 1﷯﷯ Hence we can write this as 1﷮ 𝑥﷮2﷯ + 1﷯ 𝑥 − 1﷯ 𝑥 + 1﷯﷯= 𝐴𝑥 + 𝐵﷮ 𝑥﷮2﷯ + 1﷯ + 𝐶﷮𝑥 − 1﷯ + 𝐷﷮𝑥 + 1﷯ 1﷮ 𝑥﷮2﷯ + 1﷯ 𝑥 − 1﷯ 𝑥 + 1﷯﷯ = 𝐴𝑥 + 𝐵﷯ 𝑥 − 1﷯ 𝑥 + 1﷯ + 𝐶 𝑥﷮2﷯ + 1﷯ 𝑥 + 1﷯ + 𝐷 𝑥﷮2﷯ + 1﷯ 𝑥 − 1﷯﷮ 𝑥﷮2﷯ + 1﷯ 𝑥 − 1﷯ 𝑥 + 1﷯﷯ By cancelling denominator 1= 𝐴𝑥+𝐵﷯ 𝑥−1﷯ 𝑥+1﷯+𝐶 𝑥﷮2﷯+1﷯ 𝑥+1﷯+𝐷 𝑥﷮2﷯+1﷯ 𝑥−1﷯ Putting x = 1 1= 𝐴𝑥+𝐵﷯ 𝑥−1﷯ 𝑥+1﷯+𝐶 𝑥﷮2﷯+1﷯ 𝑥+1﷯+𝐷 𝑥﷮2﷯+1﷯ 𝑥−1﷯ 1= 𝐴 1﷯+𝐵﷯ 1−1﷯ 1+1﷯+𝐶 1﷮2﷯+1﷯ 1+1﷯+𝐷 1﷮2﷯+1﷯ 1−1﷯ 1= 𝐴+𝐵﷯×0×2+𝐶 2﷯× 2﷯+𝐷 2﷯×0 1=0+𝐶×4+0 1=4𝐶 𝐶= 1﷮4﷯ Similarly putting x = −1, in (1) 1= 𝐴 −1﷯+𝐵﷯ −1−1﷯ −1+1﷯+𝐶 −1﷯﷮2﷯+1﷯ −1+1﷯+𝐷 −1﷯﷮2﷯+1﷯ −1−1﷯ 1= −𝐴+𝐵﷯ −2﷯ 0﷯+𝐶 1+1﷯ 0﷯+𝐷 1+1﷯ −2﷯ 1=0+0+𝐷×2× −2﷯ 1=−4𝐷 𝐷= −1﷮4﷯ Putting x = 0 , in (1) 1= 𝐴𝑥+𝐵﷯ 𝑥−1﷯ 𝑥+1﷯+𝐶 𝑥﷮2﷯+1﷯ 𝑥+1﷯+𝐷 𝑥﷮2﷯+1﷯ 𝑥−1﷯ 1= 𝐴×0+𝐵﷯ 0−1﷯ 0+1﷯+𝐶 0+1﷯ 0+1﷯+𝐷 0+1﷯ 0−1﷯ 1=𝐵 −1﷯ 1﷯+𝐶 1﷯ 1﷯+𝐷 1﷯ −1﷯ 1=𝐵+𝐶−𝐷 1=−𝐵+ 1﷮4﷯ − −1﷮4﷯﷯ 1=−𝐵+ 1﷮2﷯ 𝐵= 1﷮2﷯−1 𝐵= −1﷮2﷯ Putting 𝑥= 2 in (1). 1= (A𝑥+B) (𝑥−1) + C ( 𝑥﷮2﷯+1) (𝑥+1) + D ( 𝑥﷮2﷯+1) (𝑥−1) 1 = (2A + B) (2 − 1) (2 + 1) + C ( 2﷮2﷯+1) (2 +1) D ( 2﷮2﷯+1) (2− 1) 1 = (2A + B) (1) (3) + C (5) (3) + D (5) (1) 1 = 6A + 3B + 15C +5D 1 = 6A − 3﷮2﷯ + 15﷮4﷯ − 5﷮4﷯ 1 = 6A − 3﷮2﷯ + 10﷮4﷯ 1 = 6A + 1 6A = 1 − 1 6A = 0 A = 0 Therefore we can write 1﷮ 𝑥﷮2﷯ + 1﷯ 𝑥 − 1﷯ 𝑥 + 1﷯﷯= 𝐴𝑥 + 𝐵﷮ 𝑥﷮2﷯ + 1﷯ + 𝐶﷮𝑥 − 1﷯ + 𝐷﷮𝑥 + 1﷯ 1﷮ 𝑥﷮4﷯ − 1﷯= 0 × 𝑥 + − 1﷮2﷯﷯﷮ 𝑥﷮2﷯ + 1﷯ + 1﷮4﷯﷯﷮𝑥 − 1﷯ + −1﷮4﷯﷯﷮𝑥 + 1﷯ = −1﷮2 𝑥﷮2﷯ + 1﷯﷯ + 1﷮4 𝑥 − 1﷯﷯ − 1﷮4 𝑥 + 1﷯﷯ Integrating 𝑤.𝑟.𝑡.𝑥. ﷮﷮ 1﷮ 𝑥﷮4﷯ − 1﷯﷯ 𝑑𝑥= ﷮﷮ −1﷮2 𝑥﷮2﷯ + 1﷯﷯ + 1﷮4 𝑥 − 1﷯﷯ − 1﷮4 𝑥 + 1﷯﷯﷯﷯ 𝑑𝑥 = − 1﷮2﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥+ 1﷮4﷯ ﷮﷮ 1﷮𝑥 − 1﷯﷯ 𝑑𝑥− 1﷮4﷯ ﷮﷮ 1﷮𝑥 + 1﷯﷯ 𝑑𝑥 = −1﷮2﷯ tan﷮−1﷯﷮𝑥﷯+ 1﷮4﷯ log﷮ 𝑥−1﷯﷯− 1﷮4﷯log 𝑥+1﷯+ C = −1﷮2﷯ tan﷮−1﷯﷮𝑥﷯+ 1﷮4﷯ log﷮ 𝑥−1﷯−﷯log ﷯𝑥+1]+ C = −𝟏﷮𝟐﷯ 𝒕𝒂𝒏﷮−𝟏﷯﷮𝒙﷯+ 𝟏﷮𝟒﷯ 𝐥𝐨𝐠 𝒙 − 𝟏﷮𝒙 + 𝟏﷯﷯+ C